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Let $f:U\rightarrow \mathbb{R}$, where $f$ is $C^{1}$ and $U\subset \mathbb{R}^{n}$ is a open set. If $f$ has no critical points, show that $f(A)$ is a open set in $\mathbb{R}$, for all open set $A\subset U$.

My ideia is: Let $A\subset U$ arbitrary open set. For each point $z\in A$, we have $\frac{\partial f}{\partial x_{i}}(z)\neq 0$, for some $i\in \{1,...,n\},$ and usin the Implicit function theorem, $f$ converts a straight line segment parallel to the i-th axis, containing $z$ and small enough to be contained in $ A $, injecting and monotonically over a range containing $f(z)$ and contained in $f(A)$, then $f(A)$ is a open set. Since $A\subset U$ is a arbitray open set, then this holds true for any open subset cointained in $U$.

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The implicit function theorem is a particular case of the constant rank theorem, which says that if you have a function $f$ whose differential has constant rank (in your case, $rank\, Df = 1$) near a point $p$ in the domain, then there are open neighbourhoods $U$ of $p$ and $V$ of $f(p)$ and diffeomorphisms $u:T_p M \to U$, $v:T_{f(p)}\to V$ such that $f(U)\subseteq V$.

For each value $f(p)$ in $f(A)$ there is, at least, a point $p$ in $A$ where the requirements of the theorem hold. This yields an open neighbourhood $V_{f(p)}$ for each value $f(p)$ in $A$.

Thus, every value $f(p)$ in $f(A)$ has an open neighbourhood $V_{f(p)}$ that is contained in $f(A)$. So, since $f(A)$ is the union $\cup_{p\in A} V_{f(p)}$ of these open sets, $f(A)$ is itself open, as we wanted to show.

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